Helical gearing



y 15, 19-51 A. RQANO 2,553,384

* HELICAL GEARING Original Filed July 14, 1947 3 Sheets-Sheet 1 Fig.3

Aissahdro Rea-n o r) W 1 t/Ma A. ROANO 2,553,384

HELI CAL GEARING May 15, 1951 3 Sheets-Sheet 2 Original Filed July 14,1947 PO 5 a 1 c Fig. 6

C 3A6 L1 Fig.9

nltmwtoz A Zessandv'o Poano anrmm A. ROANO HELICAL GEARING May 15, 19513 Sheets-Sheet 5 Original Filed July 14, 1947 AZessandre Poano PatentedMay 15, 1951 UNITED STATES 2,553,384 iiiiLIeAL GEARING Alesse'nareRoano, Genoa, Itk'tiit v Original; application July 769, 42, .piviaea:and this appligatiori 1949, serial No. 103,702. In Italy July 8 Claims.(01. M -466) The present invention relates to gear trains; and moreparticularly single or double helical non-reciprocal gear wheels meshingbetween parallel axes, ahdis a division of prior a pueatiO h S. N.760,842, filed July 14, 1947.

An object of the present invention is an iin provement to gears whichhave already been studied and patented by the same applicant. In theseprevious patents, special helicoi'dally toothed gears, between parallelaxes; have been considered,- which, among" others-,- present thefollowing characteristics:

(1 The pitch diameters of the wheels are Substantially smaller thanWliat they ought to be in relation to the reduction ratio if they werecalculated according to the module system, but the axial pitches of theteeth of both wheels are the same, and; therefore the angles ofinclination of the helicoidal threads from which the teeth arefor'med,are different.

(2) The profile of the cross-sections of the teeth must be of thetrapezoid shape;

(3 The wheels have generally been considered of the double helicalsystem, but they might be of the single the. I

(4) The contact, between the engaged teeth of the pinion and wheel takesplace along a line W s i s at iin fih b 1 nner m said" previousfiatijitjs) fronfl' the plane] assing throughtheaxe'sfof t "ini'on anofthe wheel is inclined in no ri m finer (also by said previous"patents) in relation to said pla e, the transverse cross section of thetooth out: line of the pinion being a trapezoid shape, as saidbefore N Uw Through tests and studies it has been'asoertain'ed' that by usingdetermined forrr1s, reversf ibility or irreversibility of such gearsthey be ob-' tained; viz. itispossible to fix the limit conditions, onone side'of which the gear is reversible, and on the other side of whichthe gear is irreversible.

Indicating by p the angle of -inclination of the helicoidal threadfre'm' which are formed the pinion teeth, by {3 the angleqfinclination'oi the rili'coidal t -3mm h h theiset Q iith driven wheelare-iormed by Zpf the ang e between the two inclined; side flanks ofthetrapeze con s'tituting' the trapeze shaped outline of" the" pin? ionteeth, we have that, in order to obtain the irreversibility of the gear,the" following condi t'ions'niust be fulfilled er be tjaken int accountand they constitute" the object of the" invention:

(a) The said teeth eress seetieh outline" must ly .8, i,

have the inclinedfl anks of trapezoid section wiltn'gle p cei r l prisedv g j 39 and 5 2? for the pin eh and, accordingly; between 36 and 51 forthe wheel; I H V (b) By the same angle of inclinatio'ri a of the helicalwarm or pinion, th'e tooth' flank angle 9 of the aforesaid" piniontofoth, increases substan tie-11 proportionally to th e inerease of theangle of inclination e of the neuter worm of the teeth of the drivenwheel; I H I w I v I g (e) By the sarfie' angl ofinelinatioii e of thehelical worin oi driven w en tooth; the angl e of the efo'resaiep iohtooth section; in

ree-see with the increase of the angle of in} clination p r t helicalthe er the pinion tooth. (d) The angle of inclination of the rla'n'k orthe aforesaid sectionpf the eeee between the teeth of the drivenwheelvaries acco ding to a law arialogdtis to that afore n'tione'd for theninion t'eeth,'but' rema in'g, howevenalways just a little smaller thant eicorresponding angle of the aforesaid tooth' section of the pinion; HI

It is'obvious' new much" i'rn'portaht is the fact of having deterr'm d"the preceding conditlons;

' hears ty leaving unaltered the fund-meets characteristics r thecouple; vii: diameter of the hitch circles ineqliality; of the angles ofinl'i th rt trhen 1 fi m mfi' h eam d W .e diameter compared with thatit ought to have in the modular system fer the transmission r" reduirdheight' of teeth" and seems is pets 7 by a simple change in theiteirisfi as given un er the preeedehtiettersdo; ("5), (d) (d) to obtainat Will the reversibility of the irreversibility (if the gear.

is illustrated; known through the in the annexed sheet of at first ofall, what is already erev'ieus'ee'te ts-er the sameapieiieent win manotm e' line .Q tnelthi ess a the. exterior I, end of two consecutiveteeth of the driven W -4.7... e

F .3 repr ents 'iseqt onii qrm l to t eir):

inade according to a plane parallel to plane of Fig. 1.

Fig. is, a figure analogous to Fig. 1, predisposed for reference to thesuccessive figure.

Fig. 6 is a schematical partial plan view of the side lines of thethickness of the teeth of pinion and wheel.

Fig. 7 represents the section of a space between two teeth of the wheel,made by a plane parallel to that of Fig. 5.

Fig. 8 is a graphic representation in relation to the line of contactbetween the teeth.

Fig. 9 shows analogously to Fig. 7 the section of a space between twoteeth of the wheel.

Fig. 10 is a schematical and partial plan of the teeth of the wheel, theteeth being incline, according to the helical worms in relation to thepitch circle.

Fig. 11 is a reproduction, on a larger scale, of Figures 1 and 5 for a.geometrical demonstration. a v

Fig. 12 is a partial section of the gear, as seen normally to agenerating line and cut by a. plane passing on the line of contactbetween the pinion and wheel teeth.

With particular reference to what has been 11- lustrated in saidfigures, we have that line .ra: (Fig. 2) represents the inclination axisof the helical Worm of the wheel teeth referred to the externalcircumference of the wheel, in relation to the normal plane tothe axis;line yy indicates the inclination axis of the helical worm DI thepinion, referred to the internal diameter increased by the teeth bottomplay (or backlash); line N-l l', perpendicular to the plane of the axes,is a line serving for geometrical constructions; the distance betweenthe middle lines of the two consecutive teeth measured along the lineO'O corresponds to the axial pitch.

. With reference to Fig. 3, we have that the distance between themiddle-lines of the teeth represents the normal pitch of the wheel.

- With reference to Fig. 6, we have that the distance between theparallels EE, E-E represents, on a double scale, the thickness of theexterior portion of a pinion tooth; the distance between the parailelsA--A, A'A represents the width of the space of tooth of the wheel incorrespondence to the external circumference; line y'y indicates theaxis of inclination of the helical Worm of pinion referred to theexternal circumference of same.

. Fig. 8 represents the triangle abc in which the hypotenuse abcorresponds to the line of contact mof Fig. 5, and the cathetus be andca are respectively obtained by a parallel joining centers 0, O and aperpendicular line to it.

In Fig. 10 the teeth are inclined according to an inclination of thehelical worm on the pitch circle.

- A couple of irreversible toothed wheels, already practically built,answering satisfactorily to all requirements has the followingdimensions:

.Pim'on.Inner diameter 32.60, pitch diameter 38,

From the above data it is seen (see Figure 1) thatthe height of theteeth which take part to 4 the driving is 3.60 and that there is a playat the bottom of the spaces of 0.90.

Therefore, the inclination of the helices of the pinion related to theinner diameter increased by the said play and with regards to a plane atright angles to the axis is given by:

Ang. tan 0.1175l=6 42' 7" while the inclination of the helices of thewheel, with respect to the outer circumference and also related to theplane at right angles to the axis is Ang. tan 0.35523=l9 33 22" As theaxial pitch of the gears is 12.70, the normal pitch of the wheel isequal to 12.70 cos 19 33 22"'=11.96.

By means of the above data, Figure 2 might be drawn, which represents apartial plan view of two consecutive teeth of the wheel, in which thelines yy and new are the axes of the helices of the pinion and of thewheel, respectively, while the distance between the midlines of the twoteeth corresponds to the above determined normal pitch, viz. to 11.96.

Let us now determine the thickness of the teeth fronts:

Having drawn two consecutive teeth of the wheel, Figures 3, seen inelevation and at right angles to the helices, in which we know that thenormal pitch equals 11.96, in order to satis-, fy to the condition thatthe shape of the cross section of the sheet must be very trapezoidal,must be the angle abc'=45. Then the thickness of the fronts of the wheelteeth and that of the fronts of the pinion teeth which, as known,penetrate into the spaces between the teeth up to the depth of 3.60 are:

Ang. tan

Ang. tan

11.96- (3.60Xtan 45 X2) 11.96 7.20 2 2 As however the thickness of theteeth fronts must be obviously greater than that of the wheel teeth, letus reduce this latter to 2.16 mm. Therefore the width of the Wheelspaces at the outer circumference, seen perpendicularly to the helices,is:

while the width of the same spaces, likewise at the outer circumference,but seen at right angles to a generator, is:

9.80 cos 19 33' 22" Hence JL=%=5.l9 (see Figure 2 [Draw in Figure 2, aparallel to line y'y' at a distance from this latter of this lineintersects the edge of the gear teeth at point a. which, graphically,results at a distance of little less than 4 mm. from the plane of theaxes. From said point a draw a parallel to line 0', so as to determinethe points a, h, e, 2', Figure 2 and the point r in Figure 1.] e

We fix the particulars of heights ,andcorresponding thicknesses of thepinion teeth which will be used later on to prove that the contact.takes place only at point r situated on the outer circumference-of thewheel and at a distance of a quantity=3.'70 from the plane of axes OO.

All data reported between the above square parentheses are approximate.The same have been useful only for determining that point 1, Figure 1,comes to be approximately at a distance of about 4 mm. from the, planeof the, axes, as aforesaid, but this distance must be determined withmathematical precision, together with the remaining quotas. For thisreason. letv us determine first the heights and the correspondingthicknesses that the teeth should possess in order to obtain the contactalso at-v the right and left of the point 1*, at the arbitrary distancesgiven below, then by suitably shaping, theirv profile, we shall reducethe. contact. to a single point.

This may be. done as follows:

Before all, it is to be pointed out that my research consists in findingmathematically the length of the line ae, Figure 2 which, as it isapparent, when doubled, gives us the thickness which the pinion teethpossess in orderto obtain the contact at that point 1 which results fromthe projection of the point a in Figure 1.

Then let us drop a perpendicular from point 1", which is beingconsidered, on to line 00 and I obtain the chord half rz. Furthermore,imagine that an arbitrary number of other chord halves originating at r,be dropped always on to the outer circumference of the, wheel and bedrawn at the following distances from the plane of the axes: 4.60, 4.30,4, 3.70, 3.40:, 3.10. and"2.8.0.

At the point 1', which lies at a distance of 4.60 from the plane of theaxes, is:

By subtracting from the outer radius of, the,- pinion the abovedetermined value of or, the height of the pinion, tooth at. they point ris o,b,- tained, viz:

ang. sin-0:25.705e l-.4: 53142 If the helix of a pinion toothadvancesaxially- Ang; sin roz;=-ang. sin:

in a whole turn by 1270, it" advances through 14 53' 42"by: I

O I ll WIZJOXM =0.525=stroke he 360 Then ang. sin 0.04042=2 19' If thefictitious helix of a wheel space advances axially during a whole turnby 254, it advances through 2 19' by: 7

Ang. sin sor=ang. sin

Hence:

ah=ai hi==5.19*11534:3556 and ae=ah+he=3.556+0.525=4.081

4.081 2=8.162=thi-ckness, seen perpendicularly to a generatrix, that theteeth have at the height 2.905 in order to obtain the contact of thepoint r lying at a distance of 4.60 from the plane of the axes.

We fix after that, other. items which will be used later on in order torender evident that the contact takes place only at point 1' situated,on the outer circumference of the wheel at a distance 3.70 from theplane of the axes.

The above proceedin will be truly followed for the determination of theheights and of the corresponding thicknesses that the pinion teeth oughtto have in order to obtain the contact. inv the other points 1" thatlie, as said, at distance of4;30, 4, 3:70, 3.40, 3.10' and 2.80 from theplane ofthe axes.

In the course of the development, we will however omit. thespecification, as. this would lead Aug. sin r0z=ang. sin

hence 3 0 =0.492=stroke. he.

Then:

ang. sin 0103778=2 9 53" Ang. sin roz=ang. sin

it follows:

4.155; 2=8.3l0=thickness, seen at right angles to a generatrix, that thepinion teeth ought to have at the height 2.992 in order to obtain thecontact at the point r lying at a distance of 4.30 from the plane of theaxes.

At the point 1;, lying at a distance of 4 mm. from the plane of theaxes, is:

/m =113.729 oz=131113.79=17.271

. I v OT=JW=17J28 20.80-17.728=3.072=height of the teeth Furthermore:

Ang sin roz=ang sin 17.728 an'g. sin-0.22563=13 2' 24 hence 3600=0.460=stroke he Then ang. sin 0.03514=2 49' Ang. sin 7oz=ang. sin

it follows:

3600 =1.420=stroke hi oz=l31ll 3.739=17.261 OT=JFZWF3F6 17.653 I20.80-17.653=3.147=height of the teeth Furthermore: V

ang. sin 0.20959=12 53 Aug. sin roz=ang. sin

hence =0.426=stroke he Then 2 3.70 113.s0 V ang. sin 0.0325l= 151 46Ang. sin roz=ang. sin

it follows:

254 1 5 3b -l.314-stroke ht But (11::0" hence:

ah=ai-hi=5.191314:3376

and

ae=ah+he=3.8767+0.426=4.302

4.302x2=8.604=thickness, seen at right angles to a generatrix, that thepinion teeth ought to 5;

have at the'height 3.147 in order to obtain the contact of point r lyingat a distance of 3.70 from the plane of the axes.

At the point 1', lying ata distance of 3.40 from the plane of the axes,is:

o'z= /113.8O -3.40 =113.749 oz=131113.749=17.251

20.80-17.582=3.218=height of the teeth Furthermore:

3.40 Ang. sin ros-ang. sin

Then:

Ang. sin T0 z-ang. sin

ang. sin 0.02987=1 42' 42" it follows:

4.376 2=8.752=thickness, seen at right angles to a generatrix, that thepinion teeth ought to have at the height 3.218 in order toobtain thecontact of pointr lying at a distance of 3.40-

from the plane of the axes.

At the point r, lying at a distance 3.10 from the plane of the axes, is:

e'z=' /m=113.757 oz=131113.757=17.243

or=Jm= 17.5 19 20.80-17.519=3.281=height of the teeth Furthermore:

ang. sin 0.17695=10 11 31' Aug. sin roz=ang. sin

hence 360 =0.359=stroke he Then ang. sin 0.02724=1 33' 39" Ang. sinr0z=a,ng. sin

it follows:

ai=0" L=5.19 hence ahaihi=5.191.101=4.089

and g ae=ah+he=4.089+0.359=4.448

4.448 2=8.896=thickness, seen at right angles to a generatrix, that thepinion teeth ought to have at the height 3.281 in order to obtain thecontact at the point r lyin at a distance of 3.10 fro the plane of theaxes. I 7

At the point r, lying at a distance of 2.80 from the plane of the axes,is:

o z=dm= 113.765

oz: 131 -113.765= 17.235 r= /1 7. 2m=17.460

20.80 17.460 3.340 =height of the teeth Furthermore:

A '11 an in ng. s1 r0zg.-s 1146K- ang. sin 0. 16036= 9 13 39 hence:

12.70X9 13 39 0.325Stroke he Then:

2.80 Aug. sin r0 z ang. Sill 113-80 ang. sin 0.02460: 1 24 34 itfollows:

254 1 24 34 -OBQi-stroke h'l But ai=o" L=5.19 henceah=ai--hi='5.19-0.994 =4.196 and ae=ah+he=4.196+0.325=4.521

4.521 2=9.042=thickness, seen at right angles to a generatrix, that thepinion teeth ought to have at the height 3.340 in order to obtain thecontact of point r lying at a distance of 2.80 from the plane of theaxes.

We make the analysis of the previously determined points, in order toconfirm that the contact takes place only at point r situated on theouter circumference of the wheel which is at the distance of 3.70 fromthe plane of axes 00'.

Now, let us draw the figure of a pinion tooth, Figure 4, always seen atright angles to a generatrix by taking into account the height and thethickness thereof at the point r lying at a distance of 3.70 from theplane of the axes, in order the contact is obtained at that point. Infact in said figure, the data 3.147 and 8.604 are found, which have beenfound above just at the point r lying at a distance of 3.70 from theplane of the axes. Furthermore, by making the thickness of the toothfront equal to 2.32, the inclination of the profile is such as toexclude the contact at the right and left of said point 1'.

Let us first point out that (Figure 4) ang. tan 0.99s4.1=44 57 15" Froma drawing on a larger scale of the pinion tooth section made by a planepassing along a generating line of the pitch cylinder and taking intoaccount the different distances Aug. tan acb=ang.. tan

account that the increase in thickness takes place onboth sides of thetooth, is added to it and the width 8.120 of the horizontal section of-said cross-section of the tooth is obtained. As this width is smallerthan the thickness, as seen in a normal direction to a generating line,the pinion tooth must have at said length (2.905) in order to have thecontact in correspondence to the considered section (4.60) it maybeinferred that correspondence to this section, there is no contact.Indeed, it may be seen, that the calculated thickness 8.120 is smallerthan the precedingly calculated thickness 8.162 previously determined(line 24 of column 6).

From aforesaid table it can be seen that there is contact onlycorrespondingly to section rz=3.70 and that for the preceding andfollowing sections this contact fails.

By examining the above table from the bottom line upward, in the first(i. e. the last) line it is found the method which has been adapted fordetermining the thickness of the tooth at the maximum height.

In the second line (from below) it results that the tooth at the height3.340 has a thickness of 8.988, while in order to obtain the contact atthe point r lying at a distance of 2.80 from the plane of the axes, suchthickness should have been 9.042, just as it has been demonstrated aboveand therefore there is a difference of 0.054.

In the third line it is found that at the point 1', lying at a distanceof 3.10 from the plane of the axes, there is no contact due to adifference of 0.026.

In the fourth line, in which the point 1 results at 3.40 from the planeof the axes, there is no contact due to a difference of 0.007. p

It is apparent that in the fifth line the thickness of the tooth isequal to that indicated in Figure 4 and therefore there is a contact atthe point r lying at a distance of 3.70 from the plane of the axes.

Finally, at the points indicated in the sixth, seventh and eighth linethere is no contact, due to difference of 0.006, 0.016 and 0.042respectively.

Consequently I have shown the manner how the shape of the pinion teethshould be determined and have also demonstrated that by means of suchteeth it is possible to obtain the contact with the front edges of thewheel only at the point r, lying at a distance of 3.70 from the plane ofthe axes.

It is apparent that the shape of the spaces can be determined easily byfollowing the proceeding which has been adopted for the pinion of thereciprocal gear pair already mentioned.

Now let us first determine how the shape of the teeth of wheel 6 isdetermined and then how the contact is obtained, along the whole line m,Figure 1. A

Draw Figure 5, which is areproduction of Fig ure- 1, join the point nwith the gear centres and from the same point 11 drop a perpendicular toline 00', thus determining the chord half m1.

Figure 6 is a diagrammatically plan View of a part of Figure 5, in whichthere are indicated; the axis 11111 of a pinion tooth, which is inclinedfollowing the slant of the helix resulting from -'the-outercircumference and the'thickness of the front of the same tooth boundedby the lines EE, E'E (said front is drawn in Figure 6 on a double scale)the imaginary axis :car'of a wheel space, referred to the inclination ofthe helix at the outer circumference and the width of the same spacebounded by'the lines AA, AA', as wellas the projection of the point n onthe axis xx, on the axis w, on the construction line NN, onthe frontedge of the-pinion tooth indicated byline EE and on the front edgevof awheel tooth indicated bythe line AA, thus determining the points 2, e,h, a, B.

Due to the fact that the contact between the teethmust take place alongthe whole line nr, Figure 5, let us determine first the width of thewheel spaces, in order to obtain the contact only in the point 11 andafterwards proceed to determine the other points.

It is to be noted that the point n, which comes to be on the front edgeof the pinion teeth, in the plan view shown in Figure 6, falls on thesame front edge at the point 11. Therefore, in order that there shall becontact at that point, the

wheel spaces must have a width that is equal to the distance of thepoints a and i, multiplied Let us now bear in mind the condition thatthe contact between the teeth must take place at the same time andcontinuously along the whole line M (Fig. 1) and that the same line mustbe inclined by approximately 18 with respect to the axis and, bysatisfying said conditions it results that the point n comes to be at adistance of about 5 mm. from the plane of the axes. We fix the values ofthe depth of the spaces of the wheel and the corresponding widths, seennormally to a generating line, which will be used later on to prove thatthe contact takes place, only at point n situated on the externalcircumference of the pinion at a distance 4.80 from the plane of axesBut such a distance between the point 12 and the plane of the axis mustbe determined with mathematical precision,

which precision is obtained by determining first the various widthswhich the spaces ought to have in order to obtain the contact even atthe right and the left of the point 11., at the arbitrary distancesmentioned below and then, by suitably shaping the spaces, the contactwill be reduced to a single point n.

Be said distances the following ones: 5.70, 5.40, 5.10, 4.80, 450,420,3.90. v

At the point n, which lies at a distance 5.70 from the planeof the axes,is (Figure By substracting from the outer radius of the wheel the abovedetermined value of o'n, the depth of the wheel spaces at the point nlying at a distance of 5.70 from the plane of the axes, is obtained,viz:

It is to be noted that the helix which corresponds to the outercircumference of the pinion and which originates in the plane of theaxes, which is indicated in Figure 6 by the point 0'', during theangular shifting by the arc subtended to the angle nov advances axiallyby the distance he. r 1

At the same time the helix corresponding to the radius 0'11, of'thewheel, and which originates also in 0" in the plane of the axes e,Figure 6, during the angular shifting by the arc subtended to the anglenO'v advances axially by hi.

Let us now determine the lengths of said distances in the followingmanner:

ang. sin 0.27403: 1554'15" If the helix of a pinion tooth'advancesaxial.-

ly during a whole turn by 12.70 it advances through 15 54' 15" by O I IIi W =0.561 =stroke he Ang. sin nov=ang. sin

Then:'

ang. sin 0.05128=25621" If the helix of a Wheel tooth advances axiallyduring a whole turn by 254, it advances through 2 56 21" by Aug. sinno'v=ang.

2.672 2=5.344=width, seen at right angles to a generatrix, that wheelspaces ought to have at the depth 2.657 in order to obtain the contactat the point 11. lying at a distance of 5.70 from the plane of the axes.

We fix, by a similar procedure, other items which will be used, lateron, to render evident that the contact takes place only at point 11,situated on the external circumference of the pinion at a distance 4.80from the plane passing through 00. V 7' At the point 21., lying at adistance of 5.40 from the plane of the axes, is:

opmdaose 0v=13120.086=1l0.9l4

o'n= /m= 111.045

113.80111.045=2.755=depth of the spaces Furthermore ang. sin 0.25961= 152 48 Ang. sin n0v=ang. si

hen e ai=ae+ei=1.16+'1.436=2.596

2.596X2'=.5-.192=width:,. see, at right angles to a generatrix, that thewheel spaces ought to have at the depth 23755, in order to. obtain thecontact at. the point 12. lying at a distance of. 5.40 from the plane ofthe axes.

At the point 11., lyingat a. distance of 5.10; from the plane of theaxes is:

113.80-l.l0.952=2.848=depth of. the spaces.

Furthermore ang. sin 0.24519=14 11/35 Ang'. sin nov=ang. sin.

hence:

360 =0-.50=stroke. he

Then:

ang.. sin 0.0.4596'= 2. 38' 2" Ang. sinnou=Ang; sin

It follows:

ai==ae+ei= 1.16-1- 1358:2518

2'.5l8 2=5.036=Width, seen at right angles to a generatrix, that thewheel spaces oughttohave at the depth 2.848 in order to obtain thecontact at the point n lying at a distance of 5.1 0 from the plane ofthe axes.

At the point 11, lying at a distance- 4'.80 from the plane of the axes;is:

011-: cm =20-.23-s

113.80 1 10.865 2.935 =d'epth of' the spaces Furthermore:

ang. sin 023076=13 20 2.9"

Aug. sin n0v=ang,. sin

hence 3600 0.47 stroke he Then:

ang. sin 0104329=2 28' 51" Aug. sin na-'v=ang. si-n 2.44. 2=4.88=Width,seen at right angles. to a generatrix, that the wheel spaces ought to:have 14 at the point.- 11 lying at a.- distance of. 4.80 from the planeof the axes.

At the point 11. lying at a distance of 4.50 from the plane of the axes0v= VWW ZOBW a e: 131' -20;307'= 1101693 o"n=x/m= 110.784

113.80 l10.784=3.016=depth of the spaces Furthermore ang. sin 0.21634:12 29' 3'8" Ahg. sin nov=angz sin.

Hence 360 =0.44=stroke he Then:

ang. sin 0.04061=-2 19(39" Ang sin nov=ang. sin

2.36.4x2,=4.728=width, seen at right angles, to a generatrix.. that. thewheel spaces ought to have at the. depth 3.016 in order to obtain thecontact at the point 11 lying at a distance of: 4.50 from the: plane ofthe. axes.

At the point lying at a distance. of 4.20 from the plane of the. axes,is:

1 13.80 1 10.708 3 .092 depth of the. spaces Furthermore:

2.283 2=4.566=width, seen at right angles to a generatrix, that theWheel spaces must have at the depth 3.092 in order to obtain. theccntact at the point 12. lying at a distance of 4.20 from at the depth2.395 in order to obtain the contact the plane of the axes. I

ang. sin 0.03525=2 1 12' Aug. sin no11=ang. sin

it renews:

But v ei=hihe=1.425-0.381=1.044

hence ai=ae+ei=1.16+1044:2204

2.204 2 4.408=width, seen at right angles to 'a' generatrix that thewheel spaces ought to have -at -the depth 3.163 in order to obtain thecontact atthe point n lying at a distance of 3.90 from the plane of theaxes.

Now it must be borne inmind that the width .ofthe wheel spaces at theouter circumference, seen at right angles to a generatrix, equals 10.39(see Figures 2 and 6) and, by shaping the said wheel spaces so as toobtain the contact at the point 11. lying at-a distance of 4.80 from theplane of the axes, it results the Figure '7, in which the angle abc isequal to:

Ang. tan 2.935

ang. tan 0.9369 6=43 8' 8" In fact, in the said Figure 7, we find thedata 2.935 and 4.88 determining same and which have been previouslydetermined in correspondence 'of the point n lying at a distance of 4.80from the plane of the axes.

We make now the analysis of the points previously obtained with the aimof confirming that the contact takes place only at point 12 situated onthe external circumference of the pinion at a distance 4.80 from theplane of axes 00'.

Therefore the following table has been formulated by taking into accountthe data already determined and proceeding from the points n lying'atdifferent distances from the plane of the axes (see Figure 5 v From a.drawing at a larger scale of the section of the space between the wheelteeth, made by a plane passing along a generating line of .the pitchcylinder and taking into account the different distances of thehorizontal sections of said cross-section of the space between teeth, inrelation to plane 7 00 of the axes, it is proceeded in the followingmanner: First of all for' each horizontalsection, the chord no (forinstance 5.70) of the considered section is indicated after this, wemust consider the width, in correspondence'to the externalcircumference, of the space between two wheel teeth, as seenperpendicularly to a generating line of the pitch cylinder; after this,the product of the depth (2.657) of the space of the considered point,by the tangent of the angle (43 8 8") hereinafter determined, is deducedand we multiply by 2 in order to take into account that said space issymmetrical and the Width (5.40 3) resulting in correspondence to saidsection is obtained. By comparing this width (5.403) with the width(5.344), precedingly calculated in column 1.2 the spaces ought to havein said section, in order to perform the contact We may infer that inthis section there is no contact. The same observation may be made andthe same result will be obtained for all the preceding sections and forall the successivesections to that for which we have :4.80 which is theonly end in which there is the contact.

By reading the above table proceeding from the bottom line upward, it isfound that the last line is necessary only for determining the width ofthe bottom of the spaces.

In the second line from the bottom it is found that at the point n,lying at a distance of 3.90 from the plane of the axes, there is nocontact, as'otherwise the width of the wheel spaces, always seen atright angles to a generatrix, should have been equal to 4.408, just asit appears from the calculations already effected at that point, butthis width is actually 4.455 and therefore there is a difference of0.047.

In the third line it appears that at the point 11., lying at a distanceof 4.20 from the plane of the axes, there is no contact due to adifference of 0.020.

In the fourth line in which 1w=4.50 the difference is of 0.003 only.-

In the fifth line it results that the point n lying at a distance of4.80 from the plane of the axes, there is a contact, as the width of thespaces in that point is that indicated in Figure 7 And, at last, in thepoints n related in the sixth, seventh and eighth lines there is nocontact due to differences amounting to 0.009, 0.027 and 0.059;

. respectively.

Therefore I have demonstrated how the shaping of the wheel spaces inFigure 7 is determined when seen perpendicularly to a generatrix andfrom this, of course, it is easy to determine the shape of the teeth andI have also analytically demonstrated that by said shaping the contactis obtained only at the point n lying at a distanceof 4.80 from theplane of the axes.

But it is already known that there is a contact only at the point r.Therefore by joining the points 11. and 17 there results the line ofcontact m. straight line, as if a large number of other contact pointsbe taken along the same line, following the proceeding asset out above,and then these points be joined with one another, there results just astraight line, as shown in Figure 5, because the differences are soirrelevant that they are not appreciable, particularly for the purposeof the Said contact line may be assumed as al7 irreversibility as willbe demonstrated herein after.

Of course I do not extend my work up to determine the other said pointsof contact, and this only in order not to complicate the specificationfurther.

Briefly the pinion teeth, as well as the wheel spaces, Figure 7, areseen perpendicularly to a generatrix, While for the machining it isnecessary to determine the same figures as seen at right angles to theirhelices, as regards the steps to be followed in order to obtain saidviews, it is sufficient to proceed as has been said in my copendingapplication with regards to the reciprocal gears.

With the described gearing it is possible to obtain the irreversibilityeven with a transmission ratio of 1:1. In such a case, the diameter ofthe driven wheel is smaller than the diameter of the pinion and we havethe following values:

Pitch circle diameter of pinion, 80 mm.

Pitch circle diameter of wheel, 25 mm.

Axial pitch of teeth, 25.4 mm.

Semi-angle between the two lateral sides of a cross section of a tooth,of pinion, 50'

Semi-angle between the two lateral sides of a cross section of a tooth,of wheel, 18

Total height of teeth, 7.70 mm.

' By what has been shown, it becomes evident the importance and theeffect of the inclination given to the flanks or lateral sides of theteeth in order to obtain the irreversibility between parallel axes.Moreover, this inclination permits, between certain limits, the gradualdisplacement of the contact zone from the plane of the axis.

In order to demonstrate how the irreversibility is obtained let us notewhat follows.

Demonstration of the irreversibility By examination of the first table,it is found that at the right and left of the point r there is nocontact for such differences, that it is right to assume that themidline of the contact surface comes to lie just at 3.70 from the planeof the axes. Then, by examining the second table it is found that at thepoint 12 lying at a distance of 4.50 from the plane of the axes there isno contact for a difference of so little as 0003 While at the distanceof 5.10 this difference amounts to 0.009; therefore it can be assumedthat the midline of the contact surface comes to lie probably at thepoint n lying at a distance of 4.70 from the plane of the axes.

Of course, a more exact mathematical determination could be obtained byeifecting the operations on values which do not differ successively fromone another by 0.03, as done, but by a smaller amount. If the saiddifferences had been limited to 0.1, the data in Figure '7 depending onthe value of m: could have been determined directly from those derivedfrom-the value of nv=4.'70.

In practice, however, the proceeding which has been adopted is amplysufficient.

Anyhow, referring to the contact point r, Figure 1, it is already knownthat rz=3.70 and 02:17.261. And referring to the contact point t, Figure5, if 1w=4.70 is, with great approximation (The data 20.307 and 20.233correspond to the By drawing by means of these data a. right angletriangle it results that the line nr is inclined by 18 26' 4" withrespect to the plane of the axes.

In fact (see Figure 8).

Ang. tan abczang. tan 1:3=

ang'. tan 0.33333 18 26' 4" Of course, the length of line nr is equal tothe length of the hypothenuse, ab, viz:

nr=ab= /l +3 =3.16

It is to be noted that, as line nr is not disposed radially it liesbetween the teeth, on a plane whose inclination is different from theinclinetion of the profiles according to Figures 4 and '7. In fact, whena wheel space is seen perpendicularly to a generatrix and whenconsidering that the width of the same space at the point r is equal to1 0.39, and as'it has been assumed that 1112:4170, the width at thepoint 12 is, with a great approximation, equal to:

(The heights 4.88 and 4.73 correspond to th widths of the space at thattwo points n lying respectively at the distance of 4.80 and 4.50irom theplane of the axes. See 2nd table.) Y

Furthermore, by bearing in mind that the length of line nr=3.1 6 theFigure 9 may be drawn, in which is:

ang. tan 0.8-78l6=41-l7 17" Therefore the contact surface between theteeth lies on a plane inclined by 41 17 17" and consequently its lengthis equal to cos 41 17 17" Let us now draw Figure 10', which is adiagrammatical view of a part of the wheel, in which the teeth areinclined according to the, slant of the helices at the pitchcircumference, viz:

Ang. tan abc=ang. tan

Ang. tan =ang. tan 2.77051 =70 9 l0 and by setting P=kg. 100, is;

Furthermore: F=P tanfll) 9 ..10l'=100 2.7705 1= kg. 27.7.05- 1,

This'means that the axial thrust, viz. the stress at right angles'tothewheel faces, equals kgs. 277.05. This stress is represented in Figure 12by the forces S'Q' and S'Q'L Itis apparent that, due to the inclinationof the. surfaces of contact between the teeth, there are the twoparallelograms S, N, F, Q and S,'N", F", Q". It is apparent that whilethe resultantsSQ' and SQ" balance one another, th components S'F' andSF" produce their effects and from these also the forces Sr' and Sr arerespectively obtained. But line rSr is parallel to contact line nsr,Figure 11 and therefore in the same Figure 11 the force s'r' isrepresented by sT, while the force Sr, which is equal and opposite to8'1" is represented by the force sT';

Therefore, from the peripherical stress, from the measure or the angleat'the apex of the wheel and from the inclination of the surfaces ofcontact between the teeth it results that the gear pair issimultaneously subjected (see Fig. 11)

(1) To a peripherical stress sP tending to turn the wheel in clockwisedirection;

(2) To the force sT tending to the same purpose; and

(3) To the force, sT tending to stop the rotation of the gear.

At the same time, from Figure 12 it results:

Of course, also S'r=kg. 137.36.

Now, in Figure 11, from the forces sP and sT the resultant sR isobtained.

But sP=l00, sT=l37'.36; hence;

The effect of this resultant, with respect to the wheel only, is equalto V Now the lever arm will be determined in the following manner;

The point s come s to be at half the distance between the points n andr.

If from the same point s the perpendicular so to line co is dropped, bybearing in mind that in Figure 5 is nv'=4'.'70, rz. 3.70 I can write: 50

But by bearing in mind that, in the same Figure 5'is'0v=20.261 and02:17.261, it follows (see Fig. 11) I hence V 'o'c=13l18.761=l12.239 andV o's= /o'c +sc /112.239 +4.20 =112.317

And however:

Ang. tan soc=an'g. tan 5,

ang. tan sag; tan 0.037l2 2 8 '35 The angle Bso is equal to angle so'c.

hence ang. Es0=ang. EsB-l-ang. Bso- 18 26' 4"+--:

v l37.36 Ang. tan RsP-- 0 ang. tan L37360=53 5640" As the angle EsP=,is:

Ang. osH=90(ang. Es0+ang. Rsp) -90- (20 34 39+53 56 40")=15 28 41" Aug.the required lever arm is:

Therefore the effect of the resultant sR multiplied by the said leverarm is equal to:

SR X 0'H=169.90 29.97=kgmm. 5091.90

Consequently the lever arm on which the peripheral stress sP actsis:

But

sP o'A=100 l05.l5=kgmm. 10515 (1) Hence it results that the force sTreduces the effect of the peripheral stress from the product SPo'A=10515 to the product sR 0'H=509l.90.

This means that the force sT produces a braking effect that is equal to10515-5091.90=54-.23.10 kgmm. (2)

At the same time the resultant sR, due to its direction, produces athrust, passing through the wheel centre, this is equal to the intensityof the force sM. But, by analogy to what happens in all other Wheelsgeared between parallel axes, to force SM is opposed force sF, that isequal and in opposite direction, which tends to cause the pinion to turnin a direction that is opposite to theperipheral stress. On the contrarythe force sT,'that is equal and in opposite direction to force sT tendsto'cause the pinion to turn in the direction of the peripheral stress.

Therefore the resultant of said opposite forces, when summed to (2),will give the amount of the definitive braking.

Now:

sP=sM=sR cos ang. 0sH=169.90 cos Then I As however the angle sec isequal to the angle EsB=l8 26' 4"; it results:

But

hence I Y By subtracting the (l3) from (A) itresults:

V eop.6?:2 71 .97= 52s.66 kgmm. v,

aces-,3

21 Now is:

os= /oc +sc 18.761 +i2(l =19.22.;-

furthermore:

Hence the resultant of the forces sF and sT is a stress of kgs. 27.50acting perpendicularly to line nsr, in a direction that is opposite tothe peripheral stress sP and, is, of course, applied by a lever arm oA.

But, when bearing in mind. that o'A=1 05.l*,. it results:

By summing (2) with (3) it results:

This means that at the end the wheel is subjected simultaneously to themotive action (1) and to the braking action (4), whose ratio is:

Therefore the driving action due to the peripheral stress iscounteracted by a braking action amountingto '79 of the said motive ordriving action.

Thus the irreversibility of gears with parallel axes has beendemonstrated.

With particular reference to Figures 11 and 12 obtained by a crosssection of the gear along a plane passing on the line of contact nsr andobserving this section normally to a generating line of the pitchcylinder of the driven wheel, it can be demonstrated, analytically, theimportance of the tooth out line in order to obtain the irreversibilityof the gearing. In fact, if we reduce ang. =41 17 17" to lower value, wereduce also the component Sr which, as it. results from what has beensaid heretofore, is equal ST. Therefore, by reducing Sr, we reduce STalso, with the result that we cannot obtain the irreversibility.

This consideration allows, while maintaining unchanged all thecharacteristics of the wheel and pinion composing the couple, viz:

Inequality of the angles of inclination of the respective helical worms;

Reduced diameters of the wheels in relation to the transmission rationbased on the modular system;

Height of teeth;

Any other element we want to examine; to draw the conclusion that areversible couple becomes irreversible and vice versa by varying theinclination of the tooth outline. The inclination of the flanks of theteeth, in order to obtain the irreversibility of the couple, arecomprised, as it results from tests, in a field varying from 39 to 52for the sections of the pinion teeth and from 36 to 51 for the sectionsof the driven Wheel teeth.

An analytical demonstration of what has been aflirmed may be thefollowing one:

Indicating by:

P=the peripheric strain;

=the semiangle between the two lateral flanks of a cross section oftooth;

p=angle of inclination of the thread on the pitch diameter of the wheel;from Figure 10, We get:

F=P cotg p Uh Zn The resultant effect G of these moments on the mediumpoint of contact S, is given by a force which we may suppose actingalong the prolongation of the peripherical strain P, the moment of whichin relation to the wheel centre is moment of the force G=GI (Y) Thetotal resistant eiTect, in the direct motion, at the wheel centre, isobtained by adding the resultant moments of. the Expressions X and Yviz.

RH +GI (Z) while the motion moment due to the peripheric'al strain ismotion moment of P=PI (W) However, owing to the displacement of thecon-- tact zone and of the angle of inclination pof theflank of tooth,the efiect is that indicated by (Z).

From what has been said therefore, it results that the wheel, in thedirect motion, is submitted to the motion action expressed by Formula Wand to the resistant action expressed by Formula Z.

When the pinion is the motivating element, it must result PIZRH-I-GI.

When the pinion is no longer the motor, but becomes resistant, viz. whenthe force acting on it changes sign, viz. it becomes P, while the wheel,which was driven before, becomes the motor, and at the same timesatisfying the irreversibility of the couple, We must get Having in mindthe preceding mathematical expressions, we see that it must be PI$P/1+cotg 6 cos p sin H-I-GI by which are confirmed the previousafiirmations relating to the conditions to be satisfied for obtainingthe irreversibility of the gearing.

Besides, it must be reminded that an irreversible couple may becomereversible also on account of a more accurate finish of the cooperatingsurfaces of the wheel and pinion teeth; in this case,

, declare that what I claim is:

in order to obtain thegirreversibility, it is necessary to increase thedistance of the contact zone nsr from the plane, of the axes, when thefinish of said surfaces will be more accurate;;.whilethis distance canbe reduced when said surfaces are not accurately finished or are rough.Although, for descriptive reasons, the present invention has been basedon what has been previously described and reported by way of example,many additions and modifications could be brought in the embodiment ofthe invention, all of them being based, though, on its fundamentalconceptions resumed in the following claims.

Having now particularly described and ascertained the nature of my saidinvention and in what manner the same is to be performed, I

1-. An irreversible gear pair of which ,a first one of the pair is adriving gear and the second one ofthe pair is a driven gear, said gearshaving intermeshing helical teeth of which sections of the teeth in aplane common to both gears perpendicular to the axes thereof aretrapezoidal with side faces of the trapezoids at the intermeshing of thegears making line contact between a tooth of one gear and a tooth of theother gear, and said side face of the trapezoid of one gear making adifierent angle with a reference diameter of that gear from the anglemade by the side face of the trapezoid of the other gear with asimilarly located reference diameter of said other gear. a I H '2. Anirreversible gear pair in accordance with claim 1, wherein said ears areof different sizes diametrically and the said angle of the smaller oneofthe pair is greater than the said angle of the larger one of the pair.

. 3. ,Anirreversible gear pair in accordance with claim l, wherein saidgears are of different sizes diametrically and the said angle of thesmaller one of the pair is comprised between 39 and 52 and for thelarger gear is comprised between 36 and 51.

4. An irreversible gear pair'in accordance with claim 1 wherein saidgears are of different sizes diametrically and in any selected ,pair ofgears the said angles of the trapezoids are proportionate to said anglesof any other selected pair of gears.

-5. An irreversible gear pair in accordance with claim 1 wherein saidgears are of different sizes diametrically 'and' any selected pair 7wherein the smaller the smaller gear is made,"the greater. its saidangle of trapezoidal side face is made.

6. An irreversible gear pair in accordance with claim 1 wherein saidgears are of difierent sizes diametrically and wherein the smaller thesmaller gear is, the greater the said angle of the smaller gear, and.wherein the said angle of the larger gear varies substantially inproportion to the variation of 'said.1angle of the selected smallerdiametrically and wherein the smaller the smalleri gear is, the greaterthe said angle of said smaller gear, and wherein the said angle of thelarger gear varies substantially in proportion to the variation of saidangle 'of the selected smaller gear, and wherein the said angle of thelargerv gear is always a little smaller than the said angle of thecorresponding smaller gear.

8. An irreversible gear pair of which a first one of the pair is adriving gear and the second one of the pair is a driven gear, said gearshaving intermeshing helical teeth, ,a tooth of the driving gear having adriving side surface, and'a tooth .of the driven gear having a drivenside surface addressed toward the said driving surface, said surfaceshaving areas thereof in contact with each other in meshing engagement ofthe said teeth, and the addressed surfaces of the driven side of thedriven tooth and driving side of the driving tooth making differentangles with respectivediameters of said gears.

ALESSANDRO RoANo.

REFERENCES CITED,

The following references file of this patent: I

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